Solve for $x$ and $y$ by deriving an expression for $x$ from the second equation, and substituting it back into the first equation. $\begin{align*}8x+3y &= 1 \\ 8x+6y &= -2\end{align*}$
Answer: Begin by moving the $y$ -term in the second equation to the right side of the equation. $8x = -6y-2$ Divide both sides by $8$ to isolate $x$ $x = {-\dfrac{3}{4}y - \dfrac{1}{4}}$ Substitute this expression for $x$ in the first equation. $8({-\dfrac{3}{4}y - \dfrac{1}{4}}) + 3y = 1$ $-6y - 2 + 3y = 1$ Simplify by combining terms, then solve for $y$ $-3y - 2 = 1$ $-3y = 3$ $y = -1$ Substitute $-1$ for $y$ in the top equation. $8x+3( -1) = 1$ $8x-3 = 1$ $8x = 4$ $x = \dfrac{1}{2}$ The solution is $\enspace x = \dfrac{1}{2}, \enspace y = -1$.